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For a scalar field, ${\nabla }_{\mu }\varphi ={\partial }_{\mu }\varphi$. Apply the product rule to the stress-energy tensor, using metric compatibility to pass the covariant derivative through $g^{\mu \nu }$:The middle terms cancel because the Levi-Civita connection gives a symmetric Hessian of a scalar field. Thus, with the sign conventions of this question, the scalar stress-energy divergence identity becomesThe last equality is the Klein-Gordon equation. This stress-energy conservation follows from the field equation and metric compatibility; it does not require imposing the Einstein field equations.

Use the intended finite-dimensional irreducible highest weight module $A^{\varphi }$, with dominant integral highest weight $\varphi$. Its highest weight multiplicity is one. An arbitrary direct sum having that same largest representation weight would not in general satisfy this formula. Set $f={\pi }_{A^{\varphi }}$ and use the corrected denominator $D$ of part (a).

Restriction to each simple root ${\mathfrak{sl}}_2$ decomposes into symmetric finite representation weight strings, so $f$ is Weyl-invariant. Consequently $Df$ is Weyl-antisymmetric. We prove that no lower Weyl alternants occur, rather than assuming the desired character formula.

First establish the denominator identity itself. The finite sum $D$ has support in $\rho -Q_{+}$ and coefficient one at $\rho$. Antisymmetry kills all singular representation weight coefficients, since a reflection fixing the representation weight reverses their sign. Every remaining Weyl orbit has a unique regular dominant representative $\eta$. If it occurs in $D$, then $\eta \le \rho$. Since $\eta$ is integral and regular dominant, $\eta -\rho$ is dominant integral; it also belongs to $-Q_{+}$. A dominant vector $v=-\sum n_i{\alpha }_i$ satisfies $(v,v)=-\sum n_i(v,{\alpha }_i)\le 0$, so it must be zero. Thus the only possible orbit is $W\rho$. The leading coefficient givesFor the formal Laplacian $\mathcal{D}(e^{\lambda })=\Vert \lambda {\Vert }^2{e}^{\lambda }$, this implies $\mathcal{D}D=\Vert \rho {\Vert }^{2}D$.

Now choose root vectors dual under an invariant form, so $[E_{\alpha },E_{-\alpha }]=H_{\alpha }$ with $H_{\alpha }$ corresponding to the vector $\alpha$, and take the quadratic Casimirfor an orthonormal Cartan basis. On a highest vector its scalar is $c_{\varphi }=\Vert \varphi +\rho {\Vert }^2-\Vert \rho {\Vert }^2$, hence it has this scalar throughout the cyclic highest weight module.

Here is the computation of the weighted trace of its radial identity. For a positive root, let $X$ be the formal weighted trace of $E_{-\alpha }{E}_{\alpha }$ and $Y$ that of $E_{\alpha }{E}_{-\alpha }$. Cyclicity of trace and representation weight shifting give $Y=e^{\alpha }X$, while their difference is the weighted trace of $H_{\alpha }$, namely $(\alpha ,\nabla f)$. ThusTaking the trace of $C$ yieldsThe product expression for $D$ givesUsing $2\rho ={\sum }_{\alpha >0}\alpha$ and the product rule for the formal Laplacian therefore givesHence every representation weight in the support of $Df$ has norm $\Vert \varphi +\rho \Vert$. Its support is also contained in $\varphi +\rho -Q_{+}$, and it is antisymmetric. Any occurring regular dominant weight $\eta$ thus satisfies $\eta \le \varphi +\rho$. If the inequality were strict, thenbecause the first factor is a nonzero nonnegative sum of simple roots and the second is strictly dominant. This contradicts the equal-norm condition. Only the orbit $W(\varphi +\rho )$ remains. Its leading coefficient is one, proving

Write the geometric vector as $\mathbf{V}={\sum }_i{V}_i^{\prime }{\mathbf{e}}_i^{\prime }(t)$. A rotating orthonormal basis obeys ${\dot{\mathbf{e}}}_i^{\prime }=\mathbf{\Omega }\times {\mathbf{e}}_i^{\prime }$. The product rule gives the transport identityBoth terms must be expressed in the same basis before adding them. If $R(t)$ maps rotating components into inertial components, the precise component equation is $d(R{\mathbf{V}}^{\prime })/dt=R({\dot{\mathbf{V}}}^{\prime }+{\mathbf{\Omega }}^{\prime }\times {\mathbf{V}}^{\prime })$. The printed expression suppresses this basis conversion.

In canonical coordinates, the Poisson bracket isIn particular, $\{q_j,g\}={\partial }_{p_j}g$, $\{p_j,g\}=-{\partial }_{q_j}g$, $\{f,q_j\}=-{\partial }_{p_j}f$ and $\{f,p_j\}={\partial }_{q_j}f$. Hence $\{q_i,p_j\}={\delta }_{ij}$ and the coordinate–coordinate and momentum–momentum brackets vanish.

The chain rule along a Hamiltonian trajectory givesThe Poisson bracket of time-dependent conserved quantities follows by defining $D_t={\partial }_t+\{\cdot ,H\}$. The ordinary product rule and Jacobi identity implyTo see the Hamiltonian part explicitly, Jacobi rearranges to $\{\{f,g\},H\}=\{\{f,H\},g\}+\{f,\{g,H\}\}$. Therefore $D_{t}f=D_{t}g=0$ implies $D_t\{f,g\}=0$, even when the conserved functions depend explicitly on time.

Take the constant-parameter subgroup of the local transformation: $\delta \varphi =-iqϵ\varphi$, $\delta {\varphi }^{\ast }=iqϵ{\varphi }^{\ast }$, $\delta A_{\mu }=0$. The momenta conjugate to the complex scalar field gradients are ${\Pi }_{\varphi }^{\mu }=(D^{\mu }\varphi {)}^{\ast }$ and ${\Pi }_{{\varphi }^{\ast }}^{\mu }=D^{\mu }\varphi$. Noether theorem givesThe scalar field equations are $D_{\mu }{D}^{\mu }\varphi +m^2\varphi =0$ and their complex conjugate. Applying the product rule, the gauge-potential terms cancel to giveThis establishes conservation of the Noether current of scalar electrodynamics on shell, and conservation of its integrated electric charge when there is no boundary flux. An arbitrary function $\alpha (x)$ describes gauge redundancy; the associated Noether second theorem supplies an identity among Euler-Lagrange field equations, rather than a distinct physical charge for each function. Indeed its on-shell current can be written $j^{\mu }[\alpha ]=\alpha J^{\mu }-F^{\mu \nu }{\partial }_{\nu }\alpha =-{\partial }_{\nu }(\alpha F^{\mu \nu })$. The antisymmetric tensor makes its divergence vanish. Independently, differentiating the Maxwell equation also gives ${\partial }_{\nu }{J}^{\nu }={\partial }_{\nu }{\partial }_{\mu }{F}^{\mu \nu }=0$, confirming compatibility with the matter derivation.^[ref]

Use coordinates $x^0=t$, $x^1=x$ and first compute the mixed canonical stress-energy tensor supplied by spacetime translations:Define the energy density $\mathcal{E}={\varphi }_{1,t}{\varphi }_{2,t}+c^2{\varphi }_{1,x}{\varphi }_{2,x}$ and $\mathcal{P}={\varphi }_{1,t}{\varphi }_{2,x}+{\varphi }_{2,t}{\varphi }_{1,x}$. Direct substitution givesThe two wave equations implyFor example, differentiating $\mathcal{E}$ and replacing each ${\varphi }_{j,tt}$ by $c^2{\varphi }_{j,xx}$ gives exactly $c^2{\partial }_x\mathcal{P}$ by the product rule. These are respectively the two component equations $\boxed{{\partial }_{\mu }{T}^{\mu }{}_{\nu }=0}$, establishing stress-energy conservation explicitly.

If the given metric $g^{\mu \nu }=\mathrm{diag}(1,-1)$ is used to raise the second index in these unrescaled coordinates, the result is $T^{\mu \nu }=\left(\begin{array}{cc}\mathcal{E}& -\mathcal{P}\\ -c^2\mathcal{P}& \mathcal{E}\end{array}\right)$, still conserved. To display the symmetric relativistic form with light speed $c$, choose $X^0=ct$, $X^1=x$ instead. With ${\partial }_{X^0}=c^{-1}{\partial }_t$, the corresponding tensor isThis states the coordinate convention explicitly rather than setting $c=1$ in some components but not others.

Set $C(\beta )=[A,e^{-\beta B}]$. The product rule givesMultiplying by the integrating factor $e^{\beta B}$ and integrating yields the commutator of an operator with an exponential:Equivalently, differentiate $e^{\beta B}A{e}^{-\beta B}-A$ directly; its derivative is the integrand and it vanishes at zero. The argument is exact for finite matrices, and for bounded operators with their norm-convergent exponentials; unbounded linear operators require an appropriate common domain.

Choose $x$ vertically upward. Let $G(x,t)$ be the gas-occupied part of a pipe cross-section, of area $\alpha A$. Define the phase mean mass density, mass-weighted velocity, second moment of velocity, and phase mean pressure byThe void fraction $\alpha$ is a volume fraction, not a mass fraction. The gas momentum in a slice of length $dx$ is $A\alpha {\bar{\rho }}_g\bar{v}dx$; its advected momentum flux is $A\alpha {\bar{\rho }}_g\overline{v^2}$. The two end pressures give the net force $-A{\partial }_x(\alpha {\bar{p}}_g)dx$. Include every other vertical force on the gas, including interfacial forces, in $M_{g}dx$. Newton's second law for this fixed slice, followed by division by $Adx$, therefore givesThis is a cross-section averaged phase momentum balance. In particular, mass weighting makes the momentum and its flux exact averages; one must not replace $\overline{v^2}$ by ${\bar{v}}^2$ without an additional closure. With liquid mean velocity $\bar{u}$, constant liquid mass density ${\rho }_{\ell }$, and liquid second moment $\overline{u^2}$, the same calculation givesWith the common pressure $p={\bar{p}}_g={\bar{p}}_{\ell }$, the specified pressure contributions to $M_g,M_{\ell }$ cancel the phase-fraction derivatives from the product rule. The remaining pressure forces per unit mixture volume are $-\alpha p_x$ and $-(1-\alpha )p_x$.

Assume that the bubbles at a given height have the stated radius $a$. Their number per unit mixture volume is $N=3\alpha /(4\pi a^3)$. The drag force on a rising bubble opposes its relative velocity, so multiplication of the single-bubble law by $N$ gives the signed magnitudeGravity acts downward; the drag force on the liquid is the opposite of that on the gas. With no wall force,A bubble retains its mass in the absence of transfer between phases. Consequently $a^3{\bar{\rho }}_g=a_0^3{\rho }_0$ andFor clarity suppress bars on $p,{\rho }_g,v$ in the steady calculation. On a macroscopic variation length $\ell$, gas inertia per unit volume scales like $\alpha {\rho }_g{v}^2/\ell$, whereas the quadratic drag force scales like $\alpha {\rho }_{\ell }{v}^2/a$. Their ratio is $O[({\rho }_g/{\rho }_{\ell })(a/\ell )]$. Thus ${\rho }_g=O({\rho }_0)\ll {\rho }_{\ell }$, together with the scale separation $a\ll \ell$ inherent in this averaging model, justifies the leading-order neglect of gas inertia. Small reference density alone does not justify it in an arbitrarily short or strongly compressed region. Neglect gas weight at the same order, with the liquid fraction not vanishing. For a stationary liquid, summing the two momentum equations givesThe leading gas balance is $0=-\alpha p_x-D_v$. Substitution of the bubble-radius law yieldsThe gas continuity equation, with constant pipe area and no mass exchange, is ${\partial }_t(\alpha {\rho }_g)+{\partial }_x(\alpha {\rho }_{g}v)=0$. Hence the steady gas mass flux $J=\alpha {\rho }_{g}v$ is constant. For $0<\alpha <1$ the pressure equation selects $v>0$, andEvaluate the latter constant at a reference point ${\rho }_g={\rho }_0$, $\alpha ={\alpha }_0$ and divide. The terminal-drift density law for compressible bubbles isThe exponent $3/5$ comes from combining mass conservation with the density dependence of bubble radius; no gas equation of state was needed.

The product rule and fluctuation incompressibility give the pointwise identityAveraging therefore givesThe fluctuation term is consequently the divergence of a Reynolds stress. The eddy viscosity model represents this momentum transport through a coefficient multiplying the mean strain-rate tensor:for its shear or deviatoric part. For example, $-\rho \overline{u^{\prime }{v}^{\prime }}\simeq {\mu }_T{U}_y$ in a thin parallel shear flow. The dynamic viscosity ${\mu }_T$ is an effective flow-dependent transport coefficient, rather than a molecular material property. Large eddies carry linear momentum between regions of different mean velocity.

There is a necessary qualification to the printed tensor equation (4). Taking its trace literally gives $0=-\rho \overline{u_i^{\prime }{u}_i^{\prime }}$, which would exclude nonzero turbulence. A consistent deviatoric eddy-viscosity closure in $d$ fluctuating dimensions isThe isotropic term is absorbed into the effective pressure $P_{\mathrm{eff}}=P+2\rho e_t/d$. Here $d=2$ for a strictly two-dimensional fluctuation model. The subsequent equations use $P$ for this effective pressure, or equivalently use (4) as a shear-stress closure. This preserves the intended thin-layer model without asserting an impossible full covariance identity.

For turbulent motions of scale $l$ and velocity scale $u_l$, the estimate ${\mu }_T\sim \rho u_{l}l$ gives ${\mu }_T/\mu \sim u_{l}l/\nu$. It is much greater than one on energetic large scales and in the inertial range well above the Kolmogorov length scale. At the molecular dissipation scales it is no longer large, and molecular transport also dominates very near a smooth solid wall.

For $y_1=(1-x{)}^{-1}$, the derivatives are $y_1^{\prime }=(1-x{)}^{-2}$ and $y_1^{\prime \prime }=2(1-x{)}^{-3}$. Consequently the left-hand side of the ordinary differential equation iswhere $x\ne 1$.

Use reduction of order, writing $y=u/(1-x)$. A direct application of the product rule givesFor $v=u^{\prime }$ the resulting first-order ordinary differential equation is $v^{\prime }/v=2/(1-x)$ whenever $v\ne 0$. Integration yields $v=C(1-x{)}^{-2}$ and $u=A+B/(1-x)$; the case $v=0$ is already included. Henceon either interval $x<1$ or $x>1$. The two displayed solutions of a differential equation are linearly independent, so they span the solution space of a homogeneous linear differential equation on either interval. The point $x=1$ is a singular point of a differential equation and cannot be crossed by a nonzero member of this family that remains finite there.

The repeated root of the characteristic equation suggests factoring out $e^x$. Set $y=e^{x}u$. The product rule givesThus the ordinary differential equation reduces to $u^{\prime \prime }=1$, with general solution $u=x^2/2+Ax+B$. The first initial condition gives $B=1$. Since $y^{\prime }(0)=A+B=2$, the other initial condition gives $A=1$. ThereforeSubstitution into the reduced ordinary differential equation and the two initial conditions verifies the result.

Assume all partial derivatives needed below are continuous. The multivariable chain rule iswhere the partial derivatives of $g$ are evaluated at $(u(x,y),v(x,y))$. Apply the product rule and the multivariable chain rule once more:Clairaut's theorem gives $g_{vu}=g_{uv}$, soDifferentiating $f_x$ with respect to $y$ instead gives

For $y_1=e^x/x$, on an interval avoiding zero,Substitution into the ordinary differential equation givesTo find an independent solution, put $v=xy$. The product rule gives $v^{\prime \prime }=x{y}^{\prime \prime }+2y^{\prime }$, so the ordinary differential equation becomes $v^{\prime \prime }-v=0$. Its independent solutions are $e^x$ and $e^{-x}$. HenceThe two displayed solutions are linearly independent, since their ratio $e^{2x}$ is nonconstant on any open interval. Constants can be chosen separately on the two sides of the singular point $x=0$.

Use the product rule for the Laplace operator:Thus Laplace's equation, away from the origin, reduces to $g^{\prime \prime }+3g^{\prime }/r=0$. On the line $x=0$ the displayed expression vanishes automatically; requiring the equation on an open region gives the same radial equation by continuity from $x\ne 0$. Multiplying by $r^3$ gives $(r^3{g}^{\prime }{)}^{\prime }=0$, so $g^{\prime }=C{r}^{-3}$ and $g=A+B{r}^{-2}$. ThereforeIf extension through the origin is required, $B=0$: along the horizontal axis the second term is $B/x$. The remaining harmonic function is $Ax$.

Differentiate the product $4(x-1{)}^{-2}(x+1{)}^{-1}$ using the product rule and chain rule:Combining terms over a common denominator gives

For the suggested substitution, put $y=xv(x)$. Then the product rule gives $y^{\prime }=v+x{v}^{\prime }$, while the right side of the ordinary differential equation becomes $v+v^{-3}$. Cancelling $v$ leaves the separable differential equationIntegration yields $v^4/4=\log x+C$. The initial condition gives $v(1)=1$, so $v^4=1+4\log x$. The positive root is forced by the initial value and continuity:This is the maximal connected interval containing $1$ on which the real solution exists. At the left endpoint $y=0$, where the original ordinary differential equation is undefined; extending the fourth-root formula further left would also cease to be real.

Divide by $x^2$, which is allowed because $x>0$:The integrating factor for this linear ordinary differential equation is $e^{\int dx/x}=x$. Multiplying through makes the left side a product rule derivative:Part (b) now gives $xy=2e^{x^2}+C$. Therefore

After division by $x>0$, the equation is linear with coefficient $-1/x$ on $y$. Its integrating factor is $1/x$, and the product rule givesUsing the antiderivative from the preceding part,The arbitrary constant multiplies the homogeneous solution $x$.

The linear ordinary differential equation has differential operator $(D-3{)}^2$. Put $y=e^{3x}v$ and use the product rule twice:Substitution cancels the $v$ and $v^{\prime }$ terms, leaving $v^{\prime \prime }=4$. Two integrations give $v=2x^2+C_{1}x+C_2$. Hence the general solution isThe two arbitrary constants account for the order of the differential equation; the $x^2$ factor in the particular term reflects the repeated root $3$ of the characteristic polynomial.

On an interval with $x\ne 0$, set $y=xv$. The product rule gives $y^{\prime }=v+x{v}^{\prime }$, and substitution givesThis is a separable differential equation. Since $1+v^2>0$ for real $v$, separation loses no real constant solution:Thus the general solution on any interval avoiding zero and the poles of the tangent function isAt $x=0$ the original differential equation forces $y(0)=0$, but no real solution from this family extends over a whole interval through zero: $\log |x|+C$ encounters infinitely many poles as $x\to 0$.